`(i)` A powdered substance `(A)` on treatment with fusion mixture gives a green coloured compound (B).
`(ii)` The solution of `(B)` in boiling water on acidification with dilute `H_(2)SO_(4)` gives a pink coloured compound (C ).
`(iii)` The aqueous solution of (A) on treatment with NaOH and `Br_(2)-` water gives a compound (D).
`(iv)` A solution of (D) in conc. `HNO_(3)` on treatment with lead peroxide at boiling temperature produced a compound `(E)` which was of the same colour at that of `(C )`.
`(v)` A solutionof ( A) on treatment with a solution of barium chloride gas a white precipitate of compound (F) which was insoluble in conc. `HNO_(3)` and conc. HCl.
Consider the following statements `:`
`(I)` An ions of both `(B)` and `(C )` are diamagnetic and have tetradhedral geometry.
`(II)` Anions of both (B) and (C ) are paramagnetic and have tetrahedral geometry.
`(C )` is diamagnetic but both have same tetrahedral geometry.
(IV) Green coloured compound (B) in a neutral or acidic medium disproportionates to give (C ) and (D).
Of these select the correct one from the codes given `:`

Options

A. I and III only
B. II and III only
C. II and IV only
D. III and IV only

Answer

D

Explanation

(III)(B)`MnO_4^(2-)=[Ar]^18 3d^1` paramagnetic and tetrahedral
(C )`MnO_4^(-)=[Ar]^18 3d^0` diamagnetic and tetrahedral
(IV)`3MnO_4^(2-)+4H^(+)to2MnO_4^(-)+MnO_2 + 2H_2O`
Solution of passage :
`MnSO_4(A)+2KNO_3 + K_2CO_3 to K_2MnO_4(B)+2KNO_2 + 2CO_2 + K_2SO_4`
`MnO_4^(2-)+e^(-)overset(H^+)toMnO_4^(-)`(C )
`Mn^(2+)+2OH^(-)toMn(OH)_2 darr , Mn(OH)_2 +Br_2 +2NaOH to MnO_2(D)+2NaBr+2H_2O`
`MnO_2 + 4HNO_3 to 2Mn(NO_3)_2 + 2H_2O +O_2`
`Mn(NO_3)_2+5PbO_2 + 6HNO_3 to 2HMnO_4(E)+5Pb(NO_3)_2 + 2H_2O`
`SO_4^(2-)+Ba^(2+)toBaSO_4darr`(white)(F)